Dado un número entero N , la tarea es encontrar los números de N dígitos más pequeños y más grandes que comienzan y terminan con el dígito N.
Ejemplos:
Entrada: N = 3
Salida:
Número más pequeño = 303
Número más grande = 393
Explicación:
303 es el número de 3 dígitos más pequeño que comienza y termina en 3.
393 es el número de 3 dígitos más grande que comienza y termina en 3.
Entrada: N = 1
Salida:
Número más pequeño = 1
Número más grande = 1
Explicación:
1 es tanto el número de 1 dígito más pequeño como el más grande que comienza y termina en 1.
Enfoque:
sabemos que el número de N dígitos más grande y más pequeño es 9999…9 , donde 9 se repite N veces y 1000…. 0 , donde 0 se repite N-1 veces respectivamente.
Ahora, para obtener el número de N dígitos más pequeño y más grande que comienza y termina con N , necesitamos reemplazar el primer y el último dígito del número de N dígitos más pequeño y más grande por N .
Tenemos que cuidar el caso de la esquina, es decir, cuando N = 1 , aquí tanto el número más grande como el más pequeño serán 1 .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find n digit
// largest number starting
// and ending with n
string findNumberL(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// largest number
string result = "";
// Find the number of
// digits in number n
int length = (int)floor(log10(n) + 1);
// Append 9
for(int i = 1; i <= n - (2 * length); i++)
{
result += '9';
}
// To make it largest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = to_string(n) + result + to_string(n);
// Return the largest number
return result;
}
// Function to find n digit
// smallest number starting
// and ending with n
string findNumberS(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// smallest number
string result = "";
// Find the number of
// digits in number n
int length = (int)floor(log10(n) + 1);
for (int i = 1; i <= n - (2 * length); i++)
{
result += '0';
}
// To make it smallest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = to_string(n) + result + to_string(n);
// Return the smallest number
return result;
}
// Driver code
int main()
{
// Given number
int N = 3;
// Function call
cout << "Smallest Number = " << findNumberS(N) << endl;
cout << "Largest Number = " << findNumberL(N);
return 0;
}
// This code is contributed by divyeshrabadiya07
Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to find n digit
// largest number starting
// and ending with n
static String findNumberL(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// largest number
String result = "";
// Find the number of
// digits in number n
int length
= (int)Math.floor(
Math.log10(n) + 1);
// Append 9
for (int i = 1;
i <= n - (2 * length); i++) {
result += '9';
}
// To make it largest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = Integer.toString(n)
+ result
+ Integer.toString(n);
// Return the largest number
return result;
}
// Function to find n digit
// smallest number starting
// and ending with n
static String findNumberS(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// smallest number
String result = "";
// Find the number of
// digits in number n
int length
= (int)Math.floor(
Math.log10(n) + 1);
for (int i = 1; i <= n - (2 * length); i++) {
result += '0';
}
// To make it smallest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = Integer.toString(n)
+ result
+ Integer.toString(n);
// Return the smallest number
return result;
}
// Driver Code
public static void main(String[] args)
{
// Given Number
int N = 3;
// Function Call
System.out.println(
"Smallest Number = "
+ findNumberS(N));
System.out.print(
"Largest Number = "
+ findNumberL(N));
}
}
Python3
# Python3 program for the
# above approach
import math
# Function to find n digit
# largest number starting
#and ending with n
def findNumberL(n):
# Corner Case when n = 1
if (n == 1):
return "1"
# Result will store the
# n - 2*length(n) digit
# largest number
result = ""
# Find the number of
# digits in number n
length = math.floor(math.log10(n) + 1)
# Append 9
for i in range(1, n - (2 *
length) + 1):
result += '9'
# To make it largest n digit
# number starting and ending
# with n, we just need to
# append n at start and end
result = (str(n) + result +
str(n))
# Return the largest number
return result
# Function to find n digit
# smallest number starting
# and ending with n
def findNumberS(n):
# Corner Case when n = 1
if (n == 1):
return "1"
# Result will store the
# n - 2*length(n) digit
# smallest number
result = ""
# Find the number of
# digits in number n
length = math.floor(math.log10(n) + 1)
for i in range(1, n -
(2 * length) + 1):
result += '0'
# To make it smallest n digit
# number starting and ending
# with n, we just need to
# append n at start and end
result = (str(n) + result +
str(n))
# Return the smallest number
return result
# Driver Code
if __name__ == "__main__":
# Given Number
N = 3
# Function Call
print("Smallest Number = " + findNumberS(N))
print("Largest Number = "+ findNumberL(N))
# This code is contributed by Chitranayal
C#
// C# program for the above approach
using System;
class GFG{
// Function to find n digit
// largest number starting
// and ending with n
static String findNumberL(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// largest number
String result = "";
// Find the number of
// digits in number n
int length = (int)Math.Floor(
Math.Log10(n) + 1);
// Append 9
for(int i = 1;
i <= n - (2 * length); i++)
{
result += '9';
}
// To make it largest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = n.ToString() + result +
n.ToString();
// Return the largest number
return result;
}
// Function to find n digit
// smallest number starting
// and ending with n
static String findNumberS(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// smallest number
String result = "";
// Find the number of
// digits in number n
int length = (int)Math.Floor(
Math.Log10(n) + 1);
for (int i = 1;
i <= n - (2 * length); i++)
{
result += '0';
}
// To make it smallest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = n.ToString() + result +
n.ToString();
// Return the smallest number
return result;
}
// Driver Code
public static void Main(String[] args)
{
// Given number
int N = 3;
// Function call
Console.WriteLine("Smallest Number = " +
findNumberS(N));
Console.Write("Largest Number = " +
findNumberL(N));
}
}
// This code is contributed by Amit Katiyar
Javascript
<script>
// JavaScript program for the above approach
// Function to find n digit
// largest number starting
// and ending with n
function findNumberL(n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// largest number
let result = "";
// Find the number of
// digits in number n
let length
= Math.floor(
Math.log10(n) + 1);
// Append 9
for (let i = 1;
i <= n - (2 * length); i++) {
result += '9';
}
// To make it largest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = n.toString()
+ result
+ n.toString();
// Return the largest number
return result;
}
// Function to find n digit
// smallest number starting
// and ending with n
function findNumberS(n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// smallest number
let result = "";
// Find the number of
// digits in number n
let length
= Math.floor(
Math.log10(n) + 1);
for (let i = 1; i <= n - (2 * length); i++) {
result += '0';
}
// To make it smallest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = n.toString()
+ result
+ n.toString();
// Return the smallest number
return result;
}
// Driver Code
// Given Number
let N = 3;
// Function Call
document.write(
"Smallest Number = "
+ findNumberS(N) + "<br/>");
document.write(
"Largest Number = "
+ findNumberL(N));
</script>
Producción:
Smallest Number = 303 Largest Number = 393
Complejidad de tiempo: O(N)
Publicación traducida automáticamente
Artículo escrito por shubham prakash 1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA